# Evaluate the indefinite integral integrate of (dx/((x+1)(x^2+1)))

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You should use partial fraction decomposition to make easier the evaluation of the given integral such that:

`1/((x+1)(x^2+1)) = A/(x+1) + (Bx+C)/(x^2+1)`

`1 = Ax^2 + A + Bx^2 + Bx + Cx + C`

`1 = x^2(A+B) + x(B+C) + A + C`

Equating the coefficients of like powers yields:

`A+B = 0 => A = -B`

`B+C = 0 => B = -C => A = C`

`A+C = 1 => 2A = 1 => A = C = 1/2 => B = -1/2`

`1/((x+1)(x^2+1)) = (1/2)(1/(x+1) + (-x+1)/(x^2+1))`

Integrating both sides yields:

`int 1/((x+1)(x^2+1)) dx= (1/2)(int 1/(x+1)dx + int (-x+1)/(x^2+1) dx)`

`int 1/((x+1)(x^2+1)) dx = (1/2)(ln|x+1|- int x/(x^2+1)dx + int 1/(x^2+1)dx)`

You should use the following substitution `x^2+1 = t => 2xdx = dt => xdx = (dt)/2` `int 1/((x+1)(x^2+1)) dx = (1/2)(ln|x+1| - (1/2)int 1/t dt + arctan x) + c`

`int 1/((x+1)(x^2+1)) dx = (1/2)(ln|x+1|) - (1/4) ln|t| + arctan x + c`

`int 1/((x+1)(x^2+1)) dx = (1/2)(ln|x+1|) - (1/4) ln(x^2+1) + arctan x + c`

**Hence, evaluating the given integral yields `int 1/((x+1)(x^2+1)) dx = (1/2)(ln|x+1|) - (1/4) ln(x^2+1) + arctan x + c.` **